Evaluate the definite integral. $\int^3_{1}\big(4x^3-8x+1\big)\,dx = $
Answer: First, use the power rule: $\int^3_{1}\big(4x^3-8x+1\big)\,dx ~=~\left(x^4-4x^2+x\right)\Bigg|^3_{1}$ Second, plug in the limits of integration: $[3^4-4\cdot3^2+3]-[1^4-4\cdot1^2+1] = 48 +2 = 50$. The answer: $\int^3_{1}\big(4x^3-8x+1\big)\,dx ~=~ 50$